180 = 2x+x+21
159 = 3x
x = 53
Angle A = 53°
Angle B = 53°
Angle C = 53+21 = 74°
Answer:
V = 4.48m/s a = 1.57m/s²
Explanation:
ω₀ = 16rad/s
α₀ = 5.6rad/s²
r = 280mm = 0.28m
a = ?
v = ?
Angular velocity (ω₀) = velocity of acceleration / length of path
ω₀ = v / r
V = ω₀ * r
V = 16 * 0.28
V = 4.48m/s
Acceleration = ?
Angular acceleration α₀ = angular velocity (ω) / time take (t)
α₀ = ω / t .... equation i
But acceleration (a) = velocity (v) / time (t)
a = v / t
t = v / a
Put t = v / a into equation i
α₀ = ω / (v / a)
α₀ = ω * a / v
α * v = ω * a
a = (α * v) / ω
a = (5.6 * 4.48) / 16
a = 1.568m/s²
a = 1.57m/s²
Answer:
2361 Newtons
Explanation:
From the second Newton's law of motion;
F = ma
In this case;
we are given;
Mass as 9.5 g
Initial speed as 0 m/s
Final velocity as 650 m/s
Distance is 0.85 m
Using the equation;
V² = U² + 2as
But u = 0
v² = 2as
Therefore;
a = v² ÷ 2s
= 650² ÷ 2(0.85)
= 248,529.40 m/s²
But;
F = ma
= 0.0095 kg × 248,529.40 m/s²
= 2361 Newtons
Therefore;
The average net force required to accelerate the bullet is 2361 Newtons.
Reference point is most often given a value of zero to describe an object's position on a straight line
